Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $p = \dfrac{x - 6}{-3x^2 + 3x + 90} \times \dfrac{x^2 + 9x + 20}{-5x + 45} $
Solution: First factor out any common factors. $p = \dfrac{x - 6}{-3(x^2 - x - 30)} \times \dfrac{x^2 + 9x + 20}{-5(x - 9)} $ Then factor the quadratic expressions. $p = \dfrac {x - 6} {-3(x + 5)(x - 6)} \times \dfrac {(x + 5)(x + 4)} {-5(x - 9)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {(x - 6) \times (x + 5)(x + 4) } { -3(x + 5)(x - 6) \times -5(x - 9)} $ $p = \dfrac {(x + 5)(x + 4)(x - 6)} {15(x + 5)(x - 6)(x - 9)} $ Notice that $(x + 5)$ and $(x - 6)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {\cancel{(x + 5)}(x + 4)(x - 6)} {15\cancel{(x + 5)}(x - 6)(x - 9)} $ We are dividing by $x + 5$ , so $x + 5 \neq 0$ Therefore, $x \neq -5$ $p = \dfrac {\cancel{(x + 5)}(x + 4)\cancel{(x - 6)}} {15\cancel{(x + 5)}\cancel{(x - 6)}(x - 9)} $ We are dividing by $x - 6$ , so $x - 6 \neq 0$ Therefore, $x \neq 6$ $p = \dfrac {x + 4} {15(x - 9)} $ $ p = \dfrac{x + 4}{15(x - 9)}; x \neq -5; x \neq 6 $